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Illusion [34]
3 years ago
15

Given the QY’=4.125, what is QY?

Mathematics
1 answer:
Goryan [66]3 years ago
3 0

Answer:

Step-by-step explanation:

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Square ABCD has an area of 1600. E is the midpoint of AC and Fis the midpoint of AB.
ryzh [129]
B……….. thank me later.
4 0
2 years ago
What are the solutions to m2 – 9 = 0?
AysviL [449]

Answer:

The answer is m = 3, -3

Step-by-step explanation:

6 0
3 years ago
7 + (4 + 6) = 7+n name the propeerty
snow_tiger [21]

Answer:

n=15

Step-by-step explanation:

7+(4+6)=7+n

28+6=7+n

22=7+n

22-7=n

15=n

8 0
3 years ago
Drag the similar figure into the table.
Wittaler [7]

There wasn't no pictures nor choices so I'm afraid that I had to look it up... The first link was from Brainly and "Tetianaalnasraween05" was asking the same question.

Original Figure: 6 and 3

'Similar' Figure(s):

1.) 4 and 2

2.) 10 and 4

3.) 6 and 6

Ratio of sides are 3/2 and 6/4, 6/4 could be simplified as 3/2, so as my guess, it is, 4 and 2

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I hope this helps, as always. I wish you the best of luck and have a nice day, friend..

4 0
3 years ago
Verify csc^4x-cot^4x=2csc^2x-1
Oksanka [162]
First, factor the left side:

(csc^2(x))^2-(cot^2(x))^2

This can be simplified using x^2-y^2 = (x+y)(x-y) where the csc term is x and the cot term is y:

(csc^2(x)+cot^2(x))(csc^2(x)-cot^2(x))

One of our Pythagorean Identities states that:

csc^2(x)-cot^2(x)=1

So the left side is now:

csc^2(x)+cot^2(x)

Now we can use the same Pythagorean Identity to change the right side to:

2csc^2(x)-(csc^2(x)-cot^2(x))

All we did was express 1 in terms of csc and cot.

Distribute the negative:

2csc^2(x)-csc^2(x)+cot^2(x)

Combine like terms:

csc^2(x)+cot^2(x)

Now the left and right sides are the same.
8 0
3 years ago
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